Can I generate list [1 2 3 2 3 4 3 4 5 4 5 6 ....] without for loop?
Thanks!

 采纳的回答

m=3; % pairs
n=4; % n combinations
reshape((0:n-1)+(1:m)',1,[])

1 个评论

If you're using version prior to 2016b:
% last line would be
reshape(bsxfun(@plus,(0:n-1),(1:m)'),1,[])

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更多回答(2 个)

T=1:3*n;
Vec = floor(T/3) + mod(T-1,3) + 1;

2 个评论

The output is 1 2 4 2 3 5 3 4 6 :(
Correct answer should be Vec = floor(T/4) + mod(T-1,3) + 1.
Thanks!

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Andrei Bobrov
Andrei Bobrov 2019-9-8
编辑:Andrei Bobrov 2019-9-8
m=3;
n=4;
T = 1:m*n;
out = ceil(T/m) + mod(T-1,m);
or
out = floor(T/n) + mod(T-1,m) + 1;
or
out = repmat(1:m,1,n) + repelem(0:n-1,m);

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